Question

Let $f(x)$ be a differentiable function defined on $[0,2]$ such that $f^{\prime }(x)=f^{\prime }(2-x)$ for all $x\in (0,2)f(0)=1$ and $f(2)=e^2$. Then the value of ${\int }_0^{2}f(x)dx$ is :

• A. $1-e^2$
• B. $1+e^2$
• C. $2\left(1-e^2\right)$
• D. $2\left(1+e^2\right)$

Answer 1

Answer: (B)

$f^{\prime }(x)=f^{\prime }(2-x)$
$f(x)=-f(2-x)+c$
put $x=0$
$f^{\prime }(0)=-f^{\prime }(2)+c$
$c=f(0)+f(2)=1+e^2$
so, $f(x)+f(2-x)=1+e^2$
$I={\int }_0^{2}f(x)dx$
$I={\int }_0^{2}f(2-x)dx$
$2I={\int }_0^2(f(x)+f(2-x))dx$
$2I=\left(1+e^2\right){\int }_0^{2}dx$
$I=1+e^2$